(a) Expand \(r(r+1)(3r+4)\) to get \(3r^3 + 7r^2 + 4r\).

Use standard summation formulae:

\(\sum_{r=1}^{N} r^3 = \left( \frac{N(N+1)}{2} \right)^2, \quad \sum_{r=1}^{N} r^2 = \frac{N(N+1)(2N+1)}{6}, \quad \sum_{r=1}^{N} r = \frac{N(N+1)}{2}.\)

Substitute these into the expanded expression:

\(3 \sum_{r=1}^{N} r^3 + 7 \sum_{r=1}^{N} r^2 + 4 \sum_{r=1}^{N} r.\)

Simplify to get:

\(\frac{1}{12}N(9N^3 + 46N^2 + 75N + 38) = \frac{1}{12}N(N+1)(N+2)(9N+19).\)

(b) Express \(\frac{3r+4}{r(r+1)}\) in partial fractions:

\(\frac{3r+4}{r(r+1)} = \frac{4}{r} - \frac{1}{r+1}.\)

Substitute into the series:

\(\sum_{r=1}^{N} \left( \frac{4}{r} - \frac{1}{r+1} \right) \left( \frac{1}{4} \right)^{r+1}.\)

Write out terms and simplify using the method of differences:

\(\left( \frac{1}{4^2} \right) \left( 4 - \frac{1}{2} \right) + \left( \frac{1}{4^3} \right) \left( 4 - \frac{1}{3} \right) + \ldots + \left( \frac{1}{4^{N+1}} \right) \left( 4 - \frac{1}{N+1} \right).\)

The series telescopes to:

\(\frac{1}{4} \left( 1 - \frac{1}{N+1} \right).\)

(c) As \(N \to \infty\), the series becomes:

\(\frac{1}{4}.\)